Answer:
Step-by-step explanation:
Perimeter of a semicircle with diameter d = πd/2 + d
 
        
             
        
        
        
The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).
<h3>How to analyze quadratic equations</h3>
In this question we have a graph of a <em>quadratic</em> equation translated to another place of a <em>Cartesian</em> plane, whose form coincides with the <em>vertex</em> form of the equation of the parabola, whose form is:
g(x) = C · (x - h)² - k     (1)
Where:
- (h, k) - Vertex coordinates
- C - Vertex constant
By direct comparison we notice that (h, k) = (5, 1) and C = 1. Now we proceed to check if the points (x, y) = (2, 10) and (x, y) = (8, 10) belong to the parabola.
x = 2 
g(2) = (2 - 5)² + 1
g(2) = 10
x = 8
g(8) = (8 - 5)² + 1
g(8) = 10
The <em>quadratic</em> function g(x) = (x - 5)² + 1 passes through the points (2, 10) and (8, 10) and has a vertex at (5, 1).
To learn more on parabolae: brainly.com/question/21685473
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Answer:
Both plans would cost $100 if 6 gigabytes of data are used.
Explanation:
From the question, the system of equation are correctly represented by using small letter c to represent the total cost in dollars for both equations as already assumed in the question as follows:
c = 52 + 8d ........................... (1)
c = 82 + 3d ........................... (2)
Since c is common to both, equations (1) and (2) can therefore be equated and d solved for as follows:
52 + 8d = 82 + 3d
8d - 3d = 82 - 52
5d = 30
d = 30 / 5
d = 6
Substituting d = 6 into equation (1), we have:
c = 52 + (8 * 6)
c = 52 + 48
c = 100
Since d = 6 and c = 100, it therefore implies that both plans would cost $100 if 6 gigabytes of data are used.
 
        
             
        
        
        
Answer:
14 
Step-by-step explanation:
 
        
             
        
        
        
Answer:
Step-by-step explanation:
If you add the lengths of two sides, 12 and 14, you will get 28. Using the above theorem, the third side cannot be greater than or equal to 28. Therefore, the greatest possible whole-number length of the unknown side is 27.